The points with position vectors $10\,i + 3\,j$,$12\,i - 5\,j$,and $a\,i + 11\,j$ are collinear,if $a = $

The position vector of a point lying on the line joining the points whose position vectors are $\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is:

If $a = \hat{i} + 2 \hat{j} + 3 \hat{k}$,$b = 2 \hat{i} + 3 \hat{j} + \hat{k}$,$c = 8 \hat{i} + 13 \hat{j} + 9 \hat{k}$ and $x a + y b + z c = 0$,then $\frac{x y}{z^2} =$

If $\vec{a} = -4\hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{b} = 12\hat{i} - 6\hat{j} + 15\hat{k}$,then the vectors $\vec{a}$ and $\vec{b}$ are $.......$

If the position vectors of the vertices of a triangle are $a, b, c$,then find the sum of the vectors from the vertices to the centroid.

If $D, E$ and $F$ are respectively the mid-points of $AB, AC$ and $BC$ in $\triangle ABC$, then $\overrightarrow{BE} + \overrightarrow{AF}$ is equal to :